Question

A uniform rod of mass M and length l can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as in Fig. 8–63. The rod is held horizontally and then released. At the moment of release, determine (a) the angular acceleration of the rod, and (b) the linear acceleration of the tip of the rod. Assume that the force of gravity acts at the center of mass of the rod, as shown. [Hint: See Fig. 8–20g.]

Short answer

(a) The angular acceleration of the rod is \(\frac{{3g}}{{2l}}\).

(b) The linear acceleration of the tip of the rod is \(\frac{{3g}}{2}\).

Step-by-step solution

Identification of the given data

The mass of the rod is M.

The length of the rod is l.

(a) Relation between torque and angular acceleration

The torque,\(\tau \)and angular acceleration,\(\alpha \)are related as:

\(\tau = I\alpha \)

Here, I is the moment of inertia and is given as the product of the mass of the rigid body and the square of its distance from the axis of rotation.

\(I = m{r^2}\)

(a) Evaluation of the angular acceleration of the rod

The angular acceleration can be calculated using the relation between the torque and moment of inertia.

\(\begin{aligned}{c}\tau = I\alpha \\rF = I\alpha \end{aligned}\) … (i)

Here, the torque is provided by gravity and the moment of inertia of the rod about the center of mass is\(I = \frac{1}{3}M{l^2}\). Therefore, from equation (i), you get:

\(\begin{aligned}{c}r\left( {Mg} \right) = \left( {\frac{1}{3}M{l^2}} \right)\alpha \\\alpha = \frac{{3rg}}{{{l^2}}}\\ = \frac{{3\left( {\frac{l}{2}} \right)g}}{{{l^2}}}\\ = \frac{{3g}}{{2l}}\end{aligned}\)

(b) Evaluation of the linear acceleration of the rod

The relation between angular and linear acceleration is:

\(a = \alpha r\)

Here,\(r = l\). Therefore, the linear acceleration of the tip of the rod will be:

\(\begin{aligned}{c}a = \left( {\frac{{3g}}{{2l}}} \right)l\\ = \frac{{3g}}{2}\end{aligned}\)